As Steve Herman showed, you don't need to resort to taking partial derivatives for this equation to get the answer.  But to illustrate the method, I went through it.

Taking partial derivatives gets the same equations and the same answer in the case of this function.
To take a partial derivative with respect to one variable (of a function of more than one variable), treat all the other variables as if they were constants.

Calling the partial derivative of function f with respect to x: f'x
f'x = 2x - 6
f'y = 2y + 4
setting each to zero gives (x,y) = (3,-2)
plugging (3,-2) into the equation give Temp = -13.

To be sure this is a minimum, we need D>0 and f''xx > 0
where f''xx means the second partial derivative of F with respect to x
and D = (f''xx) * (f''yy)  -  (f''xy)^2

f''xx = f'x(2x - 6) = 2
f''yy = f'y(2y + 4) = 2
f''xy = f'y(2x - 6) = 0
So D = 4 which is greater than zero
and f''xx = 2 which is greater than zero
So it's a minimum.

(or just check some values slight smaller and larger than x=3 and y=-2 and see that the function then has a slightly larger value than -13)